No.
A simple example that doesn't quite work is the total order on $\mathbb{R}\times\mathbb{R}$ defined such that $(a,b) < (c,d)$ if $a<c$ or if $a=c$ and $b<d$. (In other words, uncountably many copies of $\mathbb{R}$, lined up end-to-end.) This is dense and clearly not order-isomorphic to the real line (extended or not), because it contains uncountably many disjoint intervals that are order-isomorphic to the real line (unlike the real line itself, which can contain only countably many of these). However, it is not Dedekind-complete per your definition, because $\{0\}\times\mathbb{R}$ is bounded above (say, by $(1,0)$), but has no least upper bound. This can be repaired by considering $\mathbb{R}\times (\mathbb{R} + \{-\infty,\infty\})$ instead, in which case the supremum of $\{0\}\times \mathbb{R}$ is just $(0,\infty)$; and you are guaranteed greatest lower bounds as well.
Snoop Dogg let's go through your list thoroughly.
𝐴 is non-empty
If $\alpha = (L_\alpha, R_{\alpha} ), $ $\beta = (L_\beta, R_{\beta} ) $, and $ \alpha ,\beta > 0^* =( \mathbb{Q^{-}} , \mathbb{Q^+}) $ then we have that there is an $ a \in L_\alpha $ and a $ b \in L_\beta $ such that $ a \not\in \mathbb{Q^{\leq 0}} $ and $ b \not\in \mathbb{Q^{-}} $ and by furthermore by the last rule, i.e.
If $𝑥\in A$, then there exists a $y\in A$ such that $y>x$,
we have that there is some $ a' \in L_\alpha $ and $ b' \in L_\beta $ such that $a' > a \geq 0 $ and $b' > b \geq 0 $; therefore $\alpha \otimes \beta $ is not empty since $0 ,a,b $ are all in $\alpha \otimes \beta $ (for example $0 < a'\cdot b'$ and $a' > 0$ and $b'>0$).
(A,B) is a non-trivial partition, i.e. $𝐴\neq \mathbb{Q}$
Since $\alpha = (L_\alpha, R_{\alpha} ), $ $\beta = (L_\beta, R_{\beta} ) $ both satisfy this rule there is some $r_\alpha$ and $r_\beta$ such that $r_\alpha > a $ and $r_\beta > b$ for all $ a \in L_\alpha , b \in L_\beta $; but then for all $c \in \alpha \otimes \beta$ we have that $c < r_\alpha \cdot r_\beta $ by the definition of $ \alpha \otimes \beta$ (let $a'\in L_\alpha ,b' \in L_\beta $ be witnesses to the definition of $ \alpha \otimes \beta$; then $c < a' \cdot b' < r_\alpha \cdot r_\beta $) so that $L_{\alpha \otimes \beta} \neq \mathbb{Q}$.
If $x,y\in \mathbb{Q}$, $x<y$, and $y \in A$ then $x \in A$
If $a \in L_\alpha ,b \ \in L_\beta $ witness the statement $c < a \cdot b $, i.e. $c \in \alpha \otimes \beta $, and let $d < c $ then by definition $d \in \alpha \otimes \beta $.
If $x \in A$, then there exists a $y\in A$ such that $y>x$
Let $a \in L_\alpha ,b \ \in L_\beta $ be witnesses to the statement $c < a \cdot b $ and since $\alpha, \beta$ are Dedekind cuts then the rule above, i.e.
If $x \in A$, then there exists a $y\in A$ such that $y>x$
so that there are some $a' \in L_\alpha ,b' \ \in L_\beta $ such that $a< a'$ and $b< b'$. Let $d = \frac{a \cdot b + a' \cdot b'}{2}$, then $ c <d < a' \cdot b'$ so that If $c \in L_{\alpha \otimes \beta} $, then there exists a $d \in L_{\alpha \otimes \beta} $ such that $d>c$ as was needed. QED
Best Answer
Dedekind cuts are used for creating reals from rational numbers, that is, axiomatically, the reals are THE Dedekind cuts of the rationals. Without the condition, however, every rational would have two representations as a Dedekind cut: one where it is added to the lower class, and another in which it is added to the upper class. Hence the condition.